cosxcos2xcos3x
傅晶屠2928求y=cosxcos2xcos4x的求导过程 -
裘俘使18224785374 ______ y=cosxcos2xcos4x dy/dx=-sinxcos2xcos4x+cosx[d(cos2xcos4x)/dx]=-sinxcos2xcos4x-2cosxsin2xcos4x-4cosxcos2xsin4x
傅晶屠2928已知f(x)=cosxcos2xcos4x,若f(a)=1/8,则角a不可能等于 -
裘俘使18224785374 ______[选项] A. 派/9 B. 2派/9 C. 2派/7 D. 4派/7
傅晶屠2928求cosxcos2xcos4x的导数值(要最简,不带有2x或4x了.) -
裘俘使18224785374 ______ (8sinxcosxcos2xcos4x/8sinx)' =(4sin2xcos2xcos4x/8sinx)' =(cos8x/8sinx)' ... 按公式求解即可
傅晶屠2928∫cosxcos2xcos3xdx -
裘俘使18224785374 ______ cosxcos2xcos3x = cos2x * cosxcos3x = cos2x * (1/2)[cos(x + 3x) + cos(x - 3x)] = cos2x * (1/2)[cos4x + cos2x] = (1/2)cos2xcos4x + (1/2)cos²2x = (1/2)(1/2)[cos(2x + 4x) + cos(2x - 4x)] + (1/2)(1/2)(1 + cos4x) = (1/4)cos6x + (1/4)cos2x + 1/4 + (1/4)cos4...
傅晶屠2928当x→0时,1 - cosxcos2xcos3x与axn是等价无穷小,求常数a,n. -
裘俘使18224785374 ______[答案] 当x→0时余弦函数在x=0的带佩亚诺余项的泰勒展开式: cosx= n+1 k=1 (−1)k−1x2k−2 (2k−2)!+o(x2n) 则当x→0时函数在x=0的带佩亚诺余项的二阶泰勒展开式分别为: cosx=1− 1 2x2+o(x2) cos(2x)=1− 1 2(2x)2+o(x2)=1-2x2+o(x2) cos(3x)=1− 1 ...
傅晶屠2928求极限的 x→∞ (1 - cosxcos2xcos3x)/(1 - cosx) -
裘俘使18224785374 ______[答案] 你确定不是x->0吗?如果是趋于无穷,基本上应该是极限不存在.
傅晶屠2928cos^2 x cos2x cosx^2 cos^3 x 求导数 要求给出sin 和前面公式一样的 -
裘俘使18224785374 ______[答案] =(cos^2 x)' cos2x cosx^2 cos^3 x +cos^2 x (cos2x)' cosx^2 cos^3 x+ cos^2 x cos2x( cosx^2 )'cos^3 x +cos^2 x cos2x cosx^2 (cos^3 x )' =-2cosxsinxcos2x cosx^2 cos^3 x-2sin2xcos^2 x cosx^2 cos^3 x -2xsinx^2cos^2 x cos2x cos^3 x -3sinxcos^2xcos^2 ...
傅晶屠2928cosxcos2x=1/2(cosx+cos3x) 这个没看懂 . -
裘俘使18224785374 ______[答案] 积化和差公式现在的教材中已经删除了,可以用下面的替代: 思路分析: 找到角 x与3x 的平均值;2x 让平均值 2x 出场,以平均值为主线;更改原来的角的样式过渡到左边的角 x,及2x, 思路启蒙于等差数列; cosx+cos3x=cos(2x-x)+cos(2x+x) =[...
傅晶屠29282cos2xcosx=cosx+cos3x为什么? -
裘俘使18224785374 ______[答案] 郭敦顒回答: 积化和差公式(3)——cosαcosβ=(1/2)[cos(α+β)+cos(α-β)], ∴2cosαcosβ=cos(α+β)+cos(α-β) 当α=2 x,β= x时,代入上式得, 2cos2xcosx= cos3x+ cosx,即2cos2xcosx=cosx+cos3x.
傅晶屠2928函数f(x)=cos2x+2cosx的最小值和最大值分别是? -
裘俘使18224785374 ______[答案] f(x)=cos2x+2cosx=2cos?x-1+2cosx=2(cos?x+cosx+1/4)-3/2=2(cosx+1/2)?-3/2因为cosx∈【-1.,1】所以当cosx=-1/2时,f(x)最小此时f(x)=-3/2当cosx=1时,f(x)最大此时f(x)=3